Let $Q $ be a matrix in $ \operatorname{GL}(2, \mathbb{Q}) $ and consider the group $G = \langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle := \langle Q^{i}(v) \mid i \in \mathbb Z, v \in \mathbb Z^{2} \rangle$ under vector addition.
Question: What are the prime index subgroups $H$ of $\langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$ such that for all $v \in H$, we have $Q(v) \in H$?
I am interested in the case when both $Q$ and $Q^{-1}$ have non-integral coefficients in their characteristic polynomial (Here are some equivalent conditions).
"Easy" case: When $Q$ (or $Q^{-1}$) is an integer matrix we can start with a subgroup $ H' $ of $\mathbb Z^2$ that has finite index $p$ and is invariant under $Q$, then $H=\langle Q^{i}( H') \mid i \in \mathbb Z \rangle$ would be a subgroup of $ G$ which has finite index $p$ and is invariant under $Q$.
One of the reasons we can do this is because when $Q$ is an integer matrix, we have $\langle Q^{i}(v)\mid i\in\mathbb Z,v\in\mathbb Z^{2}\rangle = \langle Q^{-i}(v)\mid i\in\mathbb N,v\in\mathbb Z^{2}\rangle$. However, it is quite difficult to show that this holds for the general rational matrices.
Where I get this question from: I am trying to find all the finite index subgroups of $G$ that are invariant under $Q$. Given a prime $p$ that is coprime to all the denominators of the entries in $Q$ and $Q^{-1}$, we know that $ \langle Q^{i}(p\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$ would have finite index $p^2$ and is invariant under $Q$, I was wondering if there are any other finite index invariant subgroup.
Thank you for reading. Any ideas for constructing such a subgroup would be really appreciated.
